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A small difference (Posted on 2011-02-07) Difficulty: 3 of 5
Given a set a(1), a(2), a(3)...a(n), n>1.
Let P0 be the number of permutations of n elements, in which no element a(i) is in place i.

Let P1 be the number of permutations of n elements, in which exactly one element is in its original place i.

Show that ABS(P0-P1)=1.

When is P0-P1 positive?

No Solution Yet Submitted by Ady TZIDON    
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Solution Solution Comment 2 of 2 |
Let P0n be the number of derangements of n elements and P1n be the number of

permutations in which one element has remained in its original position.

A derangement of [a(1),a(2),......,a(n)] can have in its first position any of the

n-1 elements a(2),...,a(n). Suppose that its first element is a(k). Now, a(1) will

appear either in the kth position or elsewhere.

If a(1) appears in position k

then the remaining n-2 elements a(2),a(3),...a(k-1),a(k+1),...,a(n) must be

deranged in positions 2,3,.....(k-1),(k+1),...,n which can be done in P0n-2 ways.

Otherwise, the remaining n-1 elements a(1),a(2),....a(k-1),a(k+1),...,a(n) must

be arranged in positions 2,3,....,n in such a way that none of them

remains in its original position and a(1) is not in position k. This is equivalent to

deranging [a(2),a(3),.....a(k-1),a(1),a(k+1),.....a(n)] since such a derangement

ensures that a(1) will not be in the kth position, and can be done in P0n-1 ways.

It follows that P0n = (n - 1)(P0n-2 + P0n-1), and this can be rewritten as

P0n - nP0n-1 = -(P0n-1 - [n - 1]P0n-2).                                              (1)

As Charlie observed, P1n = nP0n-1 since 1 element in its original position leaves

n - 1 elements to be deranged. So (1) can be rewritten and extended by

induction to give:   P0n - P1n = -(P0n-1 - P1n-1) = ...........= (-1)n-1(P01 - P11).

Since P01 = 0 and P11 = 1, it follows that  P0n - P1n = (-1)n

Thus, |P0n - P1n| = 1  and P0n - P1n is positive when n is even

  Posted by Harry on 2011-02-09 21:56:33
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